Long Questions (Question 12)

Question 12 (10 marks):
(a) The mass of honeydews produced in a plantation is normally distributed with a mean of 0.8 kg and a standard deviation of 0.25 kg. The honeydews are being classified into three grades A, B and C according to their masses:

Grade A > Grade B > Grade C

(i) The minimum mass of a grade A honeydew is 1.2 kg.
If a honeydew is picked at random from the plantation, find the probability that the honeydew is of grade A.

(ii) Find the minimum mass, in kg, of grade B honeydew if 20% of the honeydews are of grade C.

(b) At the Shoot the Duck game booth at an amusement park, the probability of winning is 25%.
Jacky bought tickets to play n games. The probability for Jacky to win once is 10 times the probability of losing all games.

(i) Find the value of n.

(ii) Calculate the standard deviation of the number of wins.

Solution:
μ = 0.8 kg, σ = 0.25 kg

(a)(i)
$\begin{array}{l}P\left(\text{grade }A\right)=P\left(X>1.2\right)\\ =P\left(Z>\frac{1.2-0.8}{0.25}\right)\\ =P\left(Z>1.6\right)\\ =0.0548\end{array}$

(a)(ii)
$\begin{array}{l}P\left(\text{grade C}\right)=0.2\\ P\left(X<m\right)=0.2\\ P\left(Z<\frac{m-0.8}{0.25}\right)=0.2\\ P\left(Z<-0.842\right)=0.2\\ \frac{m-0.8}{0.25}=-0.842\\ m-0.8=-0.2105\\ m=0.5895\\ \\ \text{Minimum mass of grade }B\text{ honeydew}\\ \text{is the same as the maximum mass of}\\ \text{grade }C\text{ honeydew}\text{.}\\ \\ \text{Minimum mass of grade }B=0.5895\text{ kg}\end{array}$


(b)
$\begin{array}{l}p=0.25,X=B\left(n,0.25\right)\\ P\left(X=r\right)={}^{n}C{}_r{p}^r{q}^{n-r}\\ ={}^{n}C{}_r{\left(0.25\right)}^r{\left(0.75\right)}^{n-r}\end{array}$

(b)(i)
$\begin{array}{l}P\left(X=1\right)=10P\left(X=0\right)\\ {}^{n}C{}_r{\left(0.25\right)}^1{\left(0.75\right)}^{n-r}=10\times {}^{n}C{}_0{\left(0.25\right)}^0{\left(0.75\right)}^n\\ n\times 0.25\times {\left(0.75\right)}^{n-r}=10\times 1\times 1\times {\left(0.75\right)}^n\\ 0.25n\times \frac{{\left(0.75\right)}^{n-1}}{{0.75}^n}=10\\ 0.25n\times {0.75}^{-1}=10\\ \frac{1}{4}n\left(\frac{4}{3}\right)=10\\ \frac{1}{3}n=10\\ n=30\end{array}$

(b)(ii)
$\begin{array}{l}n=30,p=0.25\\ \text{Standard deviation}\\ \text{=}\sqrt{np\left(1-p\right)}\\ =\sqrt{30\times 0.25\times 0.75}\\ =2.372\end{array}$